 Solution

WriteThe saddle equation is unchanged, soThe contour connected to the original positive-real integration path passes through $s_{+}$. Since $\Phi (s_{+})=0$, its steepest-descent branch hasFor $s=r{e}^{i\psi }$ and ${\lambda }^2=i{L}^2$,Thus the contour equation isIt approaches the origin along the positive imaginary axis, passes through $r=L$, $\psi =\pi /4$, and tends to infinity along the positive real axis. Along both ends the real part of the phase tends to $-\infty$, so the contour rises from the origin nearly vertically, bends through the first-quadrant saddle, and leaves nearly horizontally. The other saddle lies on a different constant-imaginary-phase contour and does not connect to this deformation.